126. If a + b = 1, c + d = 1 and a – b =	d	, then the value of c2 − d2 is
     	c

1. 1. a

      b

   
   
   

   2. b

      a

   
   
   

   3. 1

   
   
   

   4. –1

   
   
   

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   d	= a − b
   c

   
   

   ⇒	c	=	1	=	a + b
   	d		a − b		a − b

   
   

   ⇒	c + d	=	a + b + a − b	=	a
   	c − d		a + b − a + b		b

   
   (By componendo and dividendo)
   

   ⇒	1	=	a
   	c − d		b

   
   

   ⇒  (c – d) =	b
   	a

   
   

   ⇒  c2 – d2 = (c + d) (c – d) =	b
   	a

   Correct Option: B

   d	= a − b
   c

   
   

   ⇒	c	=	1	=	a + b
   	d		a − b		a − b

   
   

   ⇒	c + d	=	a + b + a − b	=	a
   	c − d		a + b − a + b		b

   
   (By componendo and dividendo)
   

   ⇒	1	=	a
   	c − d		b

   
   

   ⇒  (c – d) =	b
   	a

   
   

   ⇒  c2 – d2 = (c + d) (c – d) =	b
   	a

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127. If	b − c	+	a + c	+	a − b	= 1 and a – b + c ≠ 0
     	a		b		c

     then which one of the following relations is true ?

1. 1. 1	=	1	+	1
      c		a		b

   
   
   

   2. 1	=	1	+	1
      a		b		c

   
   
   

   3. 1	=	1	−	1
      b		a		c

   
   
   

   4. 1	=	1	+	1
      b		a		c

   
   
   

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   b − c	+	a + c	+	a − b	= 1
   a		b		c

   
   

   ⇒	b − c	+	a − b	+	a + c	− 1 = 0
   	a		c		b

   
   

   ⇒	b − c	+	a − b	+	a + c − b	= 0
   	a		c		b

   
   

   ⇒	c − b	+	b − a	=	a + c − b
   	a		c		b

   
   

   ⇒	c2 − bc + ab − a2	=	a + c − b
   	ac		b

   
   

   ⇒	(c2 − a2) − (bc − ab)	=	a + c − b
   	ac		b

   
   

   ⇒	(c − a) (c + a) − b(c − a)	=	a + c − b
   	ac		b

   
   

   ⇒	(c − a)(c + a − b)	=	a + c − b
   	ac		b

   
   

   ⇒	c − a	=	1
   	ac		b

   
   

   ⇒	c	−	a	=	1
   	ac		ac		b

   
   

   ⇒	1	−	1	=	1
   	a		c		b

   Correct Option: C

   b − c	+	a + c	+	a − b	= 1
   a		b		c

   
   

   ⇒	b − c	+	a − b	+	a + c	− 1 = 0
   	a		c		b

   
   

   ⇒	b − c	+	a − b	+	a + c − b	= 0
   	a		c		b

   
   

   ⇒	c − b	+	b − a	=	a + c − b
   	a		c		b

   
   

   ⇒	c2 − bc + ab − a2	=	a + c − b
   	ac		b

   
   

   ⇒	(c2 − a2) − (bc − ab)	=	a + c − b
   	ac		b

   
   

   ⇒	(c − a) (c + a) − b(c − a)	=	a + c − b
   	ac		b

   
   

   ⇒	(c − a)(c + a − b)	=	a + c − b
   	ac		b

   
   

   ⇒	c − a	=	1
   	ac		b

   
   

   ⇒	c	−	a	=	1
   	ac		ac		b

   
   

   ⇒	1	−	1	=	1
   	a		c		b

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128. If	a + b − c	=	b + c − a	=	c + a − b	and a + b + c ≠ 0, then
     	a + b		b + c		c + a

1. 1. a ≠ b ≠ c

   
   
   

   2. a = b = c

   
   
   

   3. a = b ≠ c

   
   
   

   4. a ≠ b = c

   
   
   

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   a + b − c	=	b + c − a	=	c + a − b
   a + b		b + c		c + a

   
   

   ⇒	a + b	−	c	=	b + c	−	a	=	c + a	−	b
   	a + b		a + b		b + c		b + c		c + a		c + a

   
   

   ⇒  1 −	c	= 1 −	a	= 1 −	b
   	a + b		b + c		c + a

   
   

   ⇒	c	=	a	=	b
   	a + b		b + c		c + a

   
   

   ⇒	a + b	=	b + c	=	c + a
   	c		a		b

   
   

   ⇒	a + b	+ 1 =	b + c	+ 1 =	c + a	+ 1
   	c		a		b

   
   

   ⇒	a + b + c	=	b + c + a	=	c + a + b
   	c		a		b

   
   

   ⇒	1	=	1	=	1	⇒ a = b = c
   	c		a		b

   Correct Option: B

   a + b − c	=	b + c − a	=	c + a − b
   a + b		b + c		c + a

   
   

   ⇒	a + b	−	c	=	b + c	−	a	=	c + a	−	b
   	a + b		a + b		b + c		b + c		c + a		c + a

   
   

   ⇒  1 −	c	= 1 −	a	= 1 −	b
   	a + b		b + c		c + a

   
   

   ⇒	c	=	a	=	b
   	a + b		b + c		c + a

   
   

   ⇒	a + b	=	b + c	=	c + a
   	c		a		b

   
   

   ⇒	a + b	+ 1 =	b + c	+ 1 =	c + a	+ 1
   	c		a		b

   
   

   ⇒	a + b + c	=	b + c + a	=	c + a + b
   	c		a		b

   
   

   ⇒	1	=	1	=	1	⇒ a = b = c
   	c		a		b

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129. If   x +	1	= 1 then the value of	2	= ?
     	x		x2 − x + 2

1. 1. 2

   
   
   

   2. 4

   
   
   

   3. 2

      3

   
   
   

   4. 1

   
   
   

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   x +	1	= 1
   	x

   
   ⇒  x² + 1 = x ⇒ x² – x + 1 = 0
   

   ∴	2	=	2
   	x2 – x + 2		x2 – x + 1 + 1

   
   

   =	2	= 2
   	0 + 1

   Correct Option: A

   x +	1	= 1
   	x

   
   ⇒  x² + 1 = x ⇒ x² – x + 1 = 0
   

   ∴	2	=	2
   	x2 – x + 2		x2 – x + 1 + 1

   
   

   =	2	= 2
   	0 + 1

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130. If   x =	√5 − √3	and y =	√5 + √3	then the value of	x2 + xy + y2	= ?
     	√5 + √3		√5 − √3		x2 − xy + y2

1. 1. 63

      61

   
   
   

   2. 67

      65

   
   
   

   3. 65

      63

   
   
   

   4. 69

      67

   
   
   

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   x =	√5 − √3	, y =	√5 + √3
   	√5 + √3		√5 − √3

   
   

   ∴  x + y =	√5 − √3	+	√5 + √3
   	√5 + √3		√5 − √3

   
   

   =	(√5 − √3)2 + (√5 + √3)2
   	(√5 + √3) × (√5 − √3)

   
   

   =	2[(√5)2 + (√3)2]
   	5 − 3

   
   = 5 + 3 = 8
   

   xy =	√5 − √3	×	√5 + √3	= 1
   	√5 + √3		√5 − √3

   
   

   =	x2 + xy + y2
   	x2 − xy + y2

   
   

   ∴	(x + y)2 − xy
   	(x + y)2 − 3xy

   
   

   =	(8)2 − 1
   	(8)2 − 3

   
   

   =	64 − 1
   	64 − 3

   
   

   =	63
   	61

   Correct Option: A

   x =	√5 − √3	, y =	√5 + √3
   	√5 + √3		√5 − √3

   
   

   ∴  x + y =	√5 − √3	+	√5 + √3
   	√5 + √3		√5 − √3

   
   

   =	(√5 − √3)2 + (√5 + √3)2
   	(√5 + √3) × (√5 − √3)

   
   

   =	2[(√5)2 + (√3)2]
   	5 − 3

   
   = 5 + 3 = 8
   

   xy =	√5 − √3	×	√5 + √3	= 1
   	√5 + √3		√5 − √3

   
   

   =	x2 + xy + y2
   	x2 − xy + y2

   
   

   ∴	(x + y)2 − xy
   	(x + y)2 − 3xy

   
   

   =	(8)2 − 1
   	(8)2 − 3

   
   

   =	64 − 1
   	64 − 3

   
   

   =	63
   	61

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